# Galois cohomology of separable closure

Let $$K$$ be a local field, $$K^{sep}$$ its separable closure, $$G = Gal(K^{sep}/ K)$$ the Galois group and $$C := \overline{K^{sep}}$$ the completion with respect to the induced valuation.

In his paper on $$p$$-divisible groups, Tate proves that if $$K$$ is a $$p$$-adic field, then the continuous cohomology groups $$H^{i}(G, C)$$ are one-dimensional over $$K$$ when $$i = 0, 1$$ and vanish otherwise.

Are these continuous cohomology groups known when $$K \simeq \mathbb{F}_{q}((t))$$ is non-Archimedean of equal characteristic?

• A local function field? If not, what completion do you want to take? Commented Mar 16, 2021 at 3:38
• Yes, I meant a field of formal Laurent series with the $t$-adic topology, and the completion is the $t$-adic one. Commented Mar 16, 2021 at 4:31

Yes, they are known: They vanish in degrees $$i>0$$, and for $$i=0$$ one gets the completed perfection of $$K$$.
Indeed, let $$K'$$ be the completed perfection of $$K$$. Then $$G_K=G_{K'}$$ as both perfection and completion do not change the etale site. But now $$K'$$ is already perfectoid, so the same techniques of almost mathematics that Tate uses as an intermediate step in his computation apply to prove that $$H^i(G_{K'},C)=0$$ for $$i>0$$, and $$H^0(G_{K'},C)=K'$$.
[Edit: I realize that it may be worth mentioning that after completion, $$\overline{K^{\mathrm{sep}}}$$ is perfect.]
• Thank you, this is very helpful. How do we identify the completed perfection of $K$ with a subspace of invariants in $C$? Commented Mar 16, 2021 at 15:36
• As $\overline{K^{\mathrm{sep}}}$ is complete and perfect, there is a map $K'\to C$ (and $C=\overline{K'^{\mathrm{sep}}}$). The Galois cohomology computation is a very special case of Proposition 7.13 here (which shows the stronger statement that $H^i(G_{K'},\mathcal O_C)$ is almost zero for $i>0$, and equal to $\mathcal O_{K'}$ for $i=0$). Commented Mar 16, 2021 at 15:40